pacman::p_load(caTools, ggplot2, dplyr)
D = read.csv("data/quality.csv")  # Read in dataset
set.seed(88)
split = sample.split(D$PoorCare, SplitRatio = 0.75)  # split vector
TR = subset(D, split == TRUE)
TS = subset(D, split == FALSE)
glm1 = glm(PoorCare ~ OfficeVisits + Narcotics, TR, family=binomial)
summary(glm1)

【A】從預測到決策

Fig 11.3 - 從預測到決策


【B】預測機率分佈 (DPP)

因為這個資料集很小，我們使用全部的資料來做模擬 (通常我們是使用測試資料集)

pred = predict(glm1, D, type="response")
y = D$PoorCare
data.frame(pred, y) %>% 
  ggplot(aes(x=pred, fill=factor(y))) + 
  geom_histogram(bins=20, col='white', position="stack", alpha=0.5) +
  ggtitle("Distribution of Predicted Probability (DPP,FULL)") +
  xlab("predicted probability")

【C】試算期望報酬

報酬矩陣 Payoff Matrix

payoff = matrix(c(0,-100,-10,-50),2,2)
rownames(payoff) = c("FALSE","TRUE")
colnames(payoff) = c("NoAct","Act")
payoff
      NoAct Act
FALSE     0 -10
TRUE   -100 -50

期望報酬 Expected Payoff

cutoff = seq(0, 1, 0.01)
result = sapply(cutoff, function(p) {
  cm = table(factor(y==1, c(F,T)), factor(pred>p, c(F,T))) 
  sum(cm * payoff) # sum of confusion * payoff matrix
  })
i = which.max(result)
par(cex=0.7, mar=c(4,4,3,1))
plot(cutoff, result, type='l', col='cyan', lwd=2, main=sprintf(
  "Optomal Expected Result: $%d @ %.2f",result[i],cutoff[i]))
abline(v=seq(0,1,0.1),h=seq(-6000,0,100),col='lightgray',lty=3)
points(cutoff[i], result[i], pch=20, col='red', cex=2)

🌷 因為資料點太少，以上模擬的曲線很不平整，以下我們將資料點數擴大為10K

library(MASS)

Attaching package: 'MASS'
The following object is masked from 'package:dplyr':

    select
d0 = subset(D, PoorCare==0)
A0 = mvrnorm(7480, colMeans(d0[,4:5]), cov(d0[,4:5]))
d1 = subset(D, PoorCare==1)
A1 = mvrnorm(2520, colMeans(d1[,4:5]), cov(d1[,4:5]))
A = rbind(A0, A1) %>% as.data.frame() 
A$PoorCare = c(rep(0L, 7480), rep(1L, 2520))
table(A$PoorCare)

   0    1 
7480 2520

資料框A之中有一萬個保戶的資料

y = A$PoorCare
pred = predict(glm1, A, type="response")
#save(y, pred, payoff, file="Sim11B.Rdata")
cutoff = seq(0, 1, 0.01)
result = sapply(cutoff, function(p) {
  cm = table(factor(y==1, c(F,T)), factor(pred>p, c(F,T))) 
  sum(cm * payoff) # sum of confusion * payoff matrix
  }) 
i = which.max(result)
par(cex=0.7, mar=c(4,4,3,1))
plot(cutoff, result, type='l', col='cyan', lwd=2, main=sprintf(
  "Optomal Expected Result: K$%.2f @ %.2f",result[i],cutoff[i]))
abline(v=seq(0,1,0.1),h=seq(-300000,-100000,25000),col='lightgray',lty=3)
points(cutoff[i], result[i], pch=20, col='red', cex=2)

🌷 商務資料的通常都相當大，所以模擬曲線一般都比較平整


【D】策略模擬

使用manipulate套件，我們可以在RStudio之中做動態的模擬 …

使用manipulate套件做策略模擬

library(manipulate)
manipulate({
  payoff = matrix(c(TN,FN,FP,TP),2,2)
  cutoff = seq(0, 1, 0.01)
  result = sapply(cutoff, function(p) {
    cm = table(factor(y==1, c(F,T)), factor(pred>p, c(F,T))) 
    sum(cm * payoff) # sum of confusion * payoff matrix
  }) / 1000
  i = which.max(result)
  par(cex=0.7)
  plot(cutoff, result, type='l', col='cyan', lwd=2, main=sprintf(
    "Optomal Expected Result: K$%.2f @ %.2f",result[i],cutoff[i]),
    ylim=c(-320, -120))
  abline(v=seq(0,1,0.1),h=seq(-400,-100,25),col='lightgray',lty=3)
  points(cutoff[i], result[i], pch=20, col='red', cex=2)
},
TN = slider(-100,0,   0,"TN(無行動)",step=5),
FN = slider(-100,0,-100,"FN(平均風險成本)",step=5),
FP = slider(-100,0, -10,"FP(行動成本)",step=5),
TP = slider(-100,0, -50,"TP(行動成本+較小風險成本)",step=5)
)


🗿 練習：
執行Sim12.R，先依預設的報酬矩陣回答下列問題：
  【A】 最佳臨界機率是？ 它所對應的期望報酬是多少？
  【B】 什麼都不做時，臨界機率和期望報酬各是多少？
  【C】 每位保戶都做時，臨界機率和期望報酬各是多少？
  【D】 以上哪一種做法期的望報酬比較高？
  【E】 在所有的商務情境都是這種狀況嗎？

藉由調整報酬矩陣：
  【F】 模擬出「全不做」比「全做」還要好的狀況
  【G】 並舉出一個會發生這種狀況的商務情境

有五種成本分別為$5, $10, $15, $20, $35的介入方法，它們分別可以將風險成本從$100降低到$70, $60, $50, $40, $30 …
  【H】 它們的最佳期望報酬分別是多少？
  【I】 哪一種介入方法的最佳期望報酬是最大的呢？


🌞 參考模擬程式

pacman::p_load(plotly)

PMX = list(
  c(0,-100, -5,-75), c(0,-100,-10,-70), c(0,-100,-15,-65),
  c(0,-100,-20,-60), c(0,-100,-35,-65)) %>% 
  lapply(matrix, nrow=2, ncol=2)

df = do.call(rbind, lapply(1:length(PMX), function(i) {
  r = sapply(cutoff, function(p) {
    cm = table(factor(y==1,c(F,T)), factor(pred>p,c(F,T))) 
    sum(cm * PMX[[i]]) })
  data.frame(drug=paste0("drug_",i), cutoff=cutoff, exp.payoff=r)
  }))

top_n(df,1,exp.payoff)
    drug cutoff exp.payoff
1 drug_4   0.43    -207100
ggplot(df, aes(x=cutoff, y=exp.payoff, col=drug)) +
  geom_line() + theme_bw() -> p

ggplotly(p)